The Qualifiers for SnackDown 2021 have ended. If you have solved one problem successfully you know you have qualified. In any case, you can check your email to know your official verdict from us. The participants who’ve cleared the Qualifiers can now proceed to Round 1A of the Championship that starts TODAY (21st to 23rd October)!
Some important points: Participants who finish with a rank of <= 1500 in Round 1A will qualify for the next round of Pre-Elimination. If there are a lot of ties, more than 1500 can qualify from 1A. The contestants of Round 1A who could not move to the Pre-Elimination round will be then invited and registered for Round 1B (29th to 31st October). Participants who finish with a rank of <= 1500 in Round 1B will qualify for the Pre-Elimination round.
At each stage of the contest, we will update you via emails to your registered email id. You can also reach us at help@codechef.com or snackdown@codechef.com.
All the best!
Best of luck everyone!
The qualification criteria for Round 1A has been loosened from "Participants with rank <= 1500" to "Participants who have solved at least 4 problems".
So all participants who have solved >= 4 problems are tentatively qualified for the Pre-Elimination round (pending plagiarism checks), and others can participate in Round 1B.
The ranklist will be finalized after plagiarism checks.
The solution of forth problem was out yesterday already. So, it is irrelevant
This rule might put the contestants who solve only 3 problems into a disadvantage. It's because it might be the case that after disqualifying all the contestants who do plagiarism, the rank of the contestants who solve 3 problems is <= 1500.
If that happens (very unlikely), then they will also be considered as qualified.
Were there any plagiarism checks done at all in the end? I was checking submissions of some users (1, 2) who cheated in Pre-Elimination round and they have identical solutions (1, 2) for BINFLIP in Round 1A as well! And they passed! And there are a lot of cases like that! So now I'm curious do you just not care at all or your plagiarism detection system is such a junk?
How to solve the last problem ?
Lazy Machine is similar to AGC30D — Inversion Sum
dp[i][j][l] is the Expected distance between the final positions of ith and jth no when no at ith position ends up at higher place than no at jth position and you can only make [l...k] swaps.
Then there are ways to make this DP O(n^2+nk) time and O(n^2) memory similar to the soln of "Inversion Sum" mentioned above.